Solve Convex Function: Show yf(y^{-1}\textbf{x}) is Convex | Mathmos6

[Mathmos6]

Homework Statement

How do I show that if f\in C^2 \text{(}\mathbb{R}\text{)} is convex then the function yf(y^{-1}\textbf{x}) is convex on (x,y):y>0?

Homework Equations

I know the standard definitions and whatnot about convexity, but I tried chugging through the algebra and didn't have any luck, can anyone show me a nice way to solve this?

Thanks!

-Mathmos6

 

[Mark44]

Show us what you tried...

 

[Mathmos6]

Show us what you tried...

I figured yf(y^{-1}\textbf{x})=yf(\frac{x}{y},1), so set f_x=\frac{\partial}{\partial{x}}f(\frac{x}{y},1), f_{yy}=\frac{\partial^2}{\partial{y^2}}f(\frac{x}{y},1) and so on:

then we get

\frac{\partial{}}{\partial{x}}(yf(\frac{x}{y},1))=yf_x \Rightarrow \frac{\partial{}^2}{\partial{x^2}}=yf_{xx}

and

\frac{\partial{}^2}{\partial{x}\partial{y}} (yf(\frac{x}{y},1))=f_x+yf_{xy}

Also, W.R.T y:

\frac{\partial{}}{\partial{y}}=yf_y+f(\frac{x}{y},1) \Rightarrow \frac{\partial{}^2}{\partial{y}^2} = yf_{yy}+2f_y - right?

Then looking at the Hessian, we know the first principal minor (=yf_xx) is >=0 if y is, because f is convex so its corresponding first principal minor must also be >=0. With regards to the second principal minor though, i.e. the determinant of the Hessian, we get

\frac{\partial{}^2}{\partial{x}^2} \frac{\partial{}^2}{\partial{y}^2} - (\frac{\partial{}^2}{\partial{x}\partial{y}})^2=y^2(f_{xx}f_{yy}-f_{xy}^2)+2y(f_{xx}f_y-f_{xy}f_x)-f_x^2

if my algebra is correct. We want to show this >=0 - the first term (the thing in the brackets multiplied by y²) is >=0 because it corresponds to the determinant of the Hessian for f - however, my concern is that I've gone wrong because something in the derivative of f(x/y,1) means this wouldn't work the same as for f(x,y) and so I'm not sure how to proceed... thanks :)